Question

A body of 0.25 kg executes shm given by y=0.4 sin 0.5 πt m. Calculate the amplitude angular frequency maximum velocity and maximum acceleration

Answer 1

Explanation:

w.k.t

y=A sinwt ,compare it with y=0.4 sin(0.5pi×t)

Amplitude=A=0.4m

angular frequency =w=0.5×pi=1.57/sec

now take y=0.4 sin(0.5×pi×t)

dy/dt=0.63 cos(0.5×pi×t)=V-------(1)

From above equation; max.velocity=0.63 m/s

Differentiate (1) w.r.t ,

d^2y/dt=0.99(-sin(0.5×pi×t) )

multiply and divide 0.4 in above equation

d^2y/dt=2.475(-0.4sin(0.5×pi×t) )

max.acceleration=2.475 m/s^2